Resultant of Two Cable Forces
Determine the magnitude of the resultant force F′R = F₁ − F₂ and its orientation θ, measured counterclockwise from the positive x axis. Cable 1 pulls on the ring with a force F₁ = 300 lb directed up and to the left along an 8–15–17 slope triangle, while cable 2 pulls with a force F₂ = 250 lb directed 50° below the negative x axis.
- \(\mathbf{F}_1 = 300\text{ lb}\), directed up and to the left along an 8–15–17 slope
- \(\mathbf{F}_2 = 250\text{ lb}\), directed 50° below the −x axis
The magnitude and direction (measured counterclockwise from the +x axis) of \(\mathbf{F}_R' = \mathbf{F}_1 - \mathbf{F}_2\).
Resolve \(\mathbf{F}_1\) into components. Its line of action rises 15 for every 8 it runs, on a hypotenuse of 17, angled up and to the left — so both the horizontal and vertical parts of the ratio carry a minus sign on \(x\) only:
Resolve \(\mathbf{F}_2\) into components. It sits 50° below the −x axis, so both parts are negative:
The question asks for \(\mathbf{F}_R' = \mathbf{F}_1 - \mathbf{F}_2\), not the sum — subtract each component in turn. Subtracting a negative \(F_{2x}\) and \(F_{2y}\) flips their sign, which is where a lot of the arithmetic mistakes on this exact problem come from:
The magnitude follows from Pythagoras on those two components:
And the direction from the arctangent of the components — both are positive, so the resultant sits in the first quadrant, just short of pointing straight up the \(y\) axis:
\(\mathbf{F}_R' = 456.6\text{ lb}, \quad \theta = 87.6^\circ \text{ from the } +x \text{ axis}\)
No solution steps shown yet.